Question

A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.

The director is going to choose 8 of these kittens at random to be in the commercial.
What is the probability that the director chooses 3 boy kittens and 5 girl kittens? Round your answer to three decimal places.

Answer 1

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.

Explanation:

The kittens are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.

This means that
`N = 8 + 10 = 18`

We want 3 boys, so
`k = 8`

The director is going to choose 8 of these kittens at random to be in the commercial.

This means that
`n = 8`

What is the probability that the director chooses 3 boy kittens and 5 girl kittens?

This is P(X = 3).

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 3) = h(3,18,8,8) = (C_(8,3)*C_(10,5))/(C_(18,8)) = 0.323`

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.